Graphing Quadratic Equations — Definition, Formula & Examples
Graphing quadratic equations is the process of plotting the U-shaped curve (parabola) described by an equation of the form y = ax² + bx + c. You find key features like the vertex, axis of symmetry, and intercepts, then sketch the curve through those points.
Given a quadratic function f(x) = ax² + bx + c where a ≠ 0, its graph is a parabola with vertex at (h, k) where h = −b/(2a) and k = f(h). The parabola opens upward when a > 0 and downward when a < 0, is symmetric about the vertical line x = h, and intersects the y-axis at (0, c).
Key Formula
Where:
- = Coefficient of x²; determines whether the parabola opens up (a > 0) or down (a < 0) and how wide or narrow it is
- = Coefficient of x; affects the horizontal position of the vertex
- = Constant term; equals the y-intercept of the parabola
How It Works
To graph a quadratic equation, start by finding the vertex using h = −b/(2a), then compute k = f(h). Draw the axis of symmetry at x = h. Next, find the y-intercept by evaluating f(0) = c, and find x-intercepts (if they exist) by solving ax² + bx + c = 0. Plot these key points, then choose one or two additional x-values on each side of the vertex to get a smooth curve. Finally, connect the points in a smooth U-shape opening up (if a > 0) or down (if a < 0).
Worked Example
Problem: Graph y = x² − 4x + 3.
Find the vertex: Use h = −b/(2a). Here a = 1, b = −4, so h = −(−4)/(2·1) = 2. Then k = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1. The vertex is (2, −1).
Find the y-intercept: Set x = 0: y = 0 − 0 + 3 = 3. The y-intercept is (0, 3).
Find the x-intercepts: Solve x² − 4x + 3 = 0 by factoring: (x − 1)(x − 3) = 0, so x = 1 and x = 3.
Plot and sketch: Plot the vertex (2, −1), the y-intercept (0, 3), and the x-intercepts (1, 0) and (3, 0). By symmetry about x = 2, the point (4, 3) mirrors (0, 3). Connect the points in a smooth upward-opening parabola.
Answer: The parabola has vertex (2, −1), opens upward, crosses the x-axis at (1, 0) and (3, 0), and crosses the y-axis at (0, 3).
Another Example
Problem: Graph y = −2x² + 8x − 6.
Find the vertex: Here a = −2, b = 8. Compute h = −8/(2·(−2)) = 2. Then k = −2(4) + 8(2) − 6 = −8 + 16 − 6 = 2. The vertex is (2, 2).
Determine direction: Since a = −2 < 0, the parabola opens downward. The vertex (2, 2) is the highest point.
Find x-intercepts: Solve −2x² + 8x − 6 = 0. Divide by −2: x² − 4x + 3 = 0, which factors as (x − 1)(x − 3) = 0. So x = 1 and x = 3.
Find the y-intercept and sketch: At x = 0: y = −6. Plot (0, −6), (1, 0), (2, 2), (3, 0), and the mirror point (4, −6). Draw a smooth downward-opening curve through these points.
Answer: The parabola has vertex (2, 2), opens downward, with x-intercepts at (1, 0) and (3, 0) and y-intercept at (0, −6).
Visualization
Why It Matters
Graphing quadratics is a core skill in Algebra 1 and Algebra 2 that appears on the SAT, ACT, and AP exams. Engineers and physicists use parabolas to model projectile motion — for instance, the arc of a thrown ball is a downward-opening parabola. Being comfortable with these graphs also prepares you for graphing higher-degree polynomials and other function families.
Common Mistakes
Mistake: Forgetting the negative sign in h = −b/(2a) and calculating h = b/(2a) instead.
Correction: Always include the negative: h = −b/(2a). For y = x² − 4x + 3, h = −(−4)/2 = 2, not −2.
Mistake: Assuming every parabola crosses the x-axis twice.
Correction: Check the discriminant b² − 4ac first. If it is negative, there are no real x-intercepts. If it equals zero, the parabola touches the x-axis at exactly one point (a double root).